如何在Shell脚本中删除文件名的扩展名? [英] How can I remove the extension of a filename in a shell script?

问题描述

以下代码有什么问题?

name='$filename | cut -f1 -d'.''

按原样，我得到了文字字符串$filename | cut -f1 -d'.'，但是如果我删除引号，我什么也得不到.同时，输入

As is, I get the literal string $filename | cut -f1 -d'.', but if I remove the quotes I don't get anything. Meanwhile, typing

"test.exe" | cut -f1 -d'.'

在外壳中提供了我想要的输出，test.我已经知道$filename已分配了正确的值.我想做的就是将没有扩展名的文件名分配给变量.

in a shell gives me the output I want, test. I already know $filename has been assigned the right value. What I want to do is assign to a variable the filename without the extension.

推荐答案

您应该使用

You should be using the command substitution syntax $(command) when you want to execute a command in script/command.

所以您的行应该是

name=$(echo "$filename" | cut -f 1 -d '.')

代码说明:

1. echo获取变量$filename的值并将其发送到标准输出
2. 然后我们抓取输出并将其通过管道传输到cut命令
3. cut将使用.作为将字符串切成段的定界符(也称为分隔符)，并通过-f选择要在输出中包含的段
4. 然后$()命令替换将获取输出并返回其值
5. 返回的值将分配给名为name
的变量

1. echo get the value of the variable $filename and send it to standard output
2. We then grab the output and pipe it to the cut command
3. The cut will use the . as delimiter (also known as separator) for cutting the string into segments and by -f we select which segment we want to have in output
4. Then the $() command substitution will get the output and return its value
5. The returned value will be assigned to the variable named name

请注意，这将使变量的一部分一直到第一个周期.:

Note that this gives the portion of the variable up to the first period .:

$ filename=hello.world
$ echo "$filename" | cut -f 1 -d '.'
hello
$ filename=hello.hello.hello
$ echo "$filename" | cut -f 1 -d '.'
hello
$ filename=hello
$ echo "$filename" | cut -f 1 -d '.'
hello

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